Question Number 193965 by Subhi last updated on 24/Jun/23 | ||
$${a},{b},{c}\:{are}\:{positive}\:{real}\:{numbers} \\ $$$${And} \\ $$$$\frac{\mathrm{1}}{{a}+{b}+\mathrm{1}}+\frac{\mathrm{1}}{{b}+{c}+\mathrm{1}}+\frac{\mathrm{1}}{{a}+{c}+\mathrm{1}}\geqslant\mathrm{1} \\ $$$${prove}\:{that}\:{a}+{b}+{c}\geqslant{ab}+{bc}+{ac} \\ $$ | ||
Answered by Subhi last updated on 26/Jun/23 | ||
$$\left({a}+{b}+\mathrm{1}\right)\left({a}+{b}+{c}^{\mathrm{2}} \right)\geqslant\left({a}+{b}+{c}\right)^{\mathrm{2}} \\ $$$$\frac{\mathrm{1}}{{a}+{b}+\mathrm{1}}\leqslant\frac{{a}+{b}+{c}^{\mathrm{2}} }{\left({a}+{b}+{c}\right)^{\mathrm{2}} } \\ $$$$\left({b}+{c}+\mathrm{1}\right)\left({b}+{c}+{a}^{\mathrm{2}} \right)\geqslant\left({b}+{c}+{a}\right)^{\mathrm{2}} \\ $$$$\frac{\mathrm{1}}{{b}+{c}+\mathrm{1}}\leqslant\frac{\left({b}+{c}+{a}^{\mathrm{2}} \right)}{\left({a}+{b}+{c}\right)^{\mathrm{2}} } \\ $$$$\left({a}+{c}+\mathrm{1}\right)\left({a}+{c}+{b}^{\mathrm{2}} \right)\geqslant\left({a}+{c}+{b}\right)^{\mathrm{2}} \\ $$$$\frac{\mathrm{1}}{{a}+{c}+\mathrm{1}}\leqslant\frac{\left({a}+{c}+{b}^{\mathrm{2}} \right)}{\left({a}+{b}+{c}\right)^{\mathrm{2}} } \\ $$$$\underset{{cyc}} {\sum}\frac{\mathrm{1}}{{a}+{b}+\mathrm{1}}\leqslant\frac{\mathrm{1}}{\left({a}+{b}+{c}\right)^{\mathrm{2}} }\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} +\mathrm{2}\left({a}+{b}+{c}\right)\right) \\ $$$$\frac{\mathrm{1}}{\left({a}+{b}+{c}\right)^{\mathrm{2}} }\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} +\mathrm{2}\left({a}+{b}+{c}\right)\right)\geqslant\mathrm{1} \\ $$$$\left({a}+{b}+{c}\right)^{\mathrm{2}} −\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} \right)\leqslant\mathrm{2}\left({a}+{b}+{c}\right) \\ $$$$\mathrm{2}\left({ab}+{bc}+{ac}\right)\leqslant\mathrm{2}\left({a}+{b}+{c}\right) \\ $$ | ||
Answered by Subhi last updated on 26/Jun/23 | ||
$$\left({a}+{b}+\mathrm{1}\right)\left({ac}^{\mathrm{2}} +{a}^{\mathrm{2}} {b}+{b}^{\mathrm{2}} {c}^{\mathrm{2}} \right)\geqslant\left({ac}+{ab}+{bc}\right)^{\mathrm{2}} \\ $$$$\frac{\mathrm{1}}{{a}+{b}+\mathrm{1}}\leqslant\frac{{ac}^{\mathrm{2}} +{a}^{\mathrm{2}} {b}+{b}^{\mathrm{2}} {c}^{\mathrm{2}} }{\left({ab}+{bc}+{ac}\right)^{\mathrm{2}} } \\ $$$$\left({b}+{c}+\mathrm{1}\right)\left({bc}^{\mathrm{2}} +{a}^{\mathrm{2}} {c}+{a}^{\mathrm{2}} {b}^{\mathrm{2}} \right)\geqslant\left({bc}+{ac}+{ab}\right)^{\mathrm{2}} \\ $$$$\frac{\mathrm{1}}{{b}+{c}+\mathrm{1}}\leqslant\frac{\left({bc}^{\mathrm{2}} +{a}^{\mathrm{2}} {c}+{a}^{\mathrm{2}} {b}^{\mathrm{2}} \right)}{\left({ab}+{bc}+{ac}\right)^{\mathrm{2}} } \\ $$$$\left({a}+{c}+\mathrm{1}\right)\left({ab}^{\mathrm{2}} +{b}^{\mathrm{2}} {c}+{a}^{\mathrm{2}} {c}^{\mathrm{2}} \right)\geqslant\left({ab}+{bc}+{ac}\right)^{\mathrm{2}} \\ $$$$\frac{\mathrm{1}}{{a}+{c}+\mathrm{1}}\leqslant\frac{\left({ab}^{\mathrm{2}} +{bc}^{\mathrm{2}} +{a}^{\mathrm{2}} {c}^{\mathrm{2}} \right)}{\left({ab}+{bc}+{ac}\right)^{\mathrm{2}} } \\ $$$$\underset{{cyc}} {\sum}\frac{\mathrm{1}}{{a}+{b}+\mathrm{1}}\leqslant\frac{\mathrm{1}}{\left({ab}+{bc}+{ac}\right)^{\mathrm{2}} }\left({ac}^{\mathrm{2}} +{a}^{\mathrm{2}} {b}+{b}^{\mathrm{2}} {c}^{\mathrm{2}} +{bc}^{\mathrm{2}} +{a}^{\mathrm{2}} {c}+{a}^{\mathrm{2}} {b}^{\mathrm{2}} +{ab}^{\mathrm{2}} +{bc}^{\mathrm{2}} +{a}^{\mathrm{2}} {c}^{\mathrm{2}} \right) \\ $$$$\left({a}+{b}+{c}\right)\left({ab}+{bc}+{ac}\right)={a}^{\mathrm{2}} {b}+{a}^{\mathrm{2}} {c}+{ab}^{\mathrm{2}} +{bc}^{\mathrm{2}} +{b}^{\mathrm{2}} {c}+{ac}^{\mathrm{2}} +\mathrm{3}{abc} \\ $$$${a}^{\mathrm{2}} {b}^{\mathrm{2}} +{b}^{\mathrm{2}} {c}^{\mathrm{2}} +{a}^{\mathrm{2}} {c}^{\mathrm{2}} \:¿\:\mathrm{3}{abc} \\ $$$$\underset{{cyc}} {\sum}\frac{\mathrm{1}}{\mathrm{2}}{a}^{\mathrm{2}} {b}^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{2}}{a}^{\mathrm{2}} {b}^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{2}}{b}^{\mathrm{2}} {c}^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{2}}{b}^{\mathrm{2}} {c}^{\mathrm{2}} \geqslant\mathrm{4}^{\mathrm{4}} \sqrt{\frac{\left({abc}\right)^{\mathrm{4}} }{\mathrm{2}^{\mathrm{4}} }}=\mathrm{2}{abc} \\ $$$$\mathrm{2}\left({a}^{\mathrm{2}} {b}^{\mathrm{2}} +{b}^{\mathrm{2}} {c}^{\mathrm{2}} +{a}^{\mathrm{2}} {c}^{\mathrm{2}} \right)\:\geqslant\:\mathrm{6}{abc} \\ $$$$\therefore\:{a}^{\mathrm{2}} {b}^{\mathrm{2}} +{b}^{\mathrm{2}} {c}^{\mathrm{2}} +{a}^{\mathrm{2}} {c}^{\mathrm{2}} \:\geqslant\:\mathrm{3}{abc} \\ $$$$\therefore\:\:\frac{\mathrm{1}}{\left({ab}+{bc}+{ac}\right)^{\mathrm{2}} }\left({a}+{b}+{c}\right)\left({ab}+{bc}+{ac}\right)\geqslant\mathrm{1} \\ $$$${a}+{b}+{c}\geqslant{ab}+{bc}+{ac} \\ $$ | ||